Social choice theory = preference aggregation = voting assuming agents tell the truth about their preferences Tuomas Sandholm Professor Computer Science Department Carnegie Mellon University

Social choice Collectively choosing among outcomes E.g. presidents Outcome can also be a vector E.g. allocation of money, goods, tasks, and resources Agents have preferences over outcomes Center knows each agent’s preferences Or agents reveal them truthfully by assumption Social choice function aggregates those preferences & picks outcome Outcome is enforced on all agents CS applications: Multiagent planning [Ephrati&Rosenschein], computerized elections [Cranor&Cytron], accepting a joint project, collaborative filtering, rating Web articles [Avery,Resnick&Zeckhauser], rating CDs...

Condorcet paradox [year 1785] Majority rule Three voters: x > z > y y > x >z z > y > x x > z > y > x Unlike in the example above, under some preferences there is a Condorcet winner, i.e., a candidate who would win a two-candidate election against each of the other candidates.

Agenda paradox x y z x y z x y z Binary protocol (majority rule) = cup Three types of agents: x > z > y (35%) y > x > z (33%) z > y > x (32%) x y z x y z x y z Power of agenda setter (e.g. chairman) Vulnerable to irrelevant alternatives (z) Plurality protocol For each agent, most preferred outcome gets 1 vote Would result in x

Pareto dominated winner paradox Voters: x > y > b > a a > x > y > b b > a > x > y

Inverted-order paradox Borda rule with 4 alternatives Each agent gives 4 points to best option, 3 to second best... Agents: x=22, a=17, b=16, c=15 Remove x: c=15, b=14, a=13 x > c > b > a a > x > c > b b > a > x > c

Borda rule also vulnerable to irrelevant alternatives Three types of agents: Borda winner is x Remove z: Borda winner is y x > z > y (35%) y > x > z (33%) z > y > x (32%)

Majority-winner paradox Agents: Majority rule with any binary protocol: a Borda protocol: b=16, a=15, c=11 a > b > c b > c > a b > a > c c > a > b

Is there a desirable way to aggregate agents’ preferences? Set of alternatives A Each agent i in {1,..,n} has a ranking <i of A Social welfare function F: Ln -> L To avoid unilluminating technicalities in proof, assume <i and < are strict total orders Some possible (weak) desiderata of F 1. Unanimity: If all voters have the same ranking, then the aggregate ranking equals that. Formally, for all < in L, F(< ,…,<) =<. 2. Nondictatorship: No voter is a dictator. Voter i is a dictator if for all <1 ,…,<n , F(<1 ,…,<n) = <i 3. Independence of irrelevant alternatives: The social preference between any alternatives a and b only depends on the voters’ preferences between a and b. Formally, for every a, b in A and every <1 ,…,<n , < ’1 ,…,< ’n in L , if we denote < = F(<1 ,…,<n) and < ’ = F(< ’1 ,…,< ’n), then a <i b <=> a < ’i b for all i implies that a < b <=> a < ’ b. Arrow’s impossibility theorem [1951]: If |A| ≥ 3, then no F satisfies desiderata 1-3.

Proof (follows logic of Nisan’s proof in Algorithmic Game Theory book, but fixes errors and includes missing part) Assume F satisfies unanimity and independence of irrelevant alternatives Lemma. Any function F: Ln -> L that satisfies unanimity and independence of irrelevant alternatives also satisfies pairwise unanimity. That is, if for all i, a <i b, then a < b. Proof. Let <* in L be such that a <* b. By unanimity, a < b in F(<* ,..,<*). If <1,...,<n are all such that a <i b, then we have a <i b <=> a <* b, and so by independence of irrelevant alternatives, for <' = F(<1,...,<n), we have a <' b. ■ Lemma (pairwise neutrality). Let >1 ,…,>n and >’1 ,…, >’n be two player profiles such that a >i b <=> c >’i d. Then a > b <=> c >’ d. Proof. Assume wlog that a > b and c not equal to d. We merge each >i and >’i into a single preference >i by putting c just above a (unless c = a) and d just below b (unless d = b) and preserving the internal order within the pairs (a,b) and (c,d). By pairwise unanimity, c > a and b > d. Thus, by transitivity, c > d. ■ Take any a not equal to b in A, and for every i in {0,…,n} define a preference profile Pi in which exactly the first i players rank b above a. By pairwise unanimity, in F(P0) we have a > b, while in F(Pn) we have b > a. Thus, for some i* the ranking of a and b flips: in F(Pi*-1) we have a > b, while in F(Pi*) we have b > a. We conclude the proof by showing that i* is a dictator: Lemma. Take any c not equal to d in A. If c >i* d then c > d. Proof. Take some alternative e that is different from c and d. For i < i* move e to the top in >i, for i > i* move e to the bottom in >i, and for i* move e so that c >i* e >i* d. By independence of irrelevant alternatives, we have not changed the social ranking between c and d. Notice that the players’ preferences for the ordered pair (d,e) are identical to their preferences for (a,b) in Pi*, but the preferences for (c,e) are identical to the preferences for (a,b) in Pi*-1, and thus using the pairwise neutrality claim, socially e > d and c > e, and thus by transitivity c > d in the preferences where e was moved. By independence of irrelevant alternatives, moving e does not affect the relative ranking of c and d; thus c > d also under the original preferences. ■ ■

Stronger version of Arrow’s theorem In Arrow’s theorem, social welfare function F outputs a ranking of the outcomes The impossibility holds even if only the highest ranked outcome is sought: Thm. Let |A| ≥ 3. If a social choice function f: Ln -> A is monotonic and Paretian, then f is dictatorial. Definition. f is monotonic if [ x = f(>) and x maintains its position in >’ ] => f(>’) = x Definition. x maintains its position whenever x >i y => x >i’ y Proof. From f we construct a social welfare function F that satisfies the conditions of Arrow’s theorem For each pair x, y of outcomes in turn, to determine whether x > y in F, move x and y to the top of each voter’s preference order don’t change their relative order (order of other alternatives is arbitrary) Lemma 1. If any two preference profiles >’ and >’’ are constructed from a preference profile > by moving some set X of outcomes to the top in this way, then f(>’) = f(>’’) Proof. Because f is Paretian, f(>’)  X. Thus f(>’) maintains its position in going from >’ to >’’. Then, by monotonicity of f, we have f(>’) = f(>’’) Note: Because f is Paretian, we have f = x or f = y (and, by lemma 1, not both) F is transitive (total order) (we omit proving this part) F is Paretian (if everyone prefers x over y, then x gets chosen and vice versa) F satisfies independence of irrelevant alternatives (immediate from lemma 1) By earlier version of the impossibility, F (and thus f) must be dictatorial. ■

Voting rules that avoid Arrow’s impossibility (by changing what the voters can express) Approval voting Each voter gets to specify which alternatives he/she approves The alternative with the largest number of approvals wins Avoids Arrow’s impossibility Unanimity Nondictatorial Independent of irrelevant alternatives Range voting Instead of submitting a ranking of the alternatives, each voter gets to assign a value (from a given range) to each alternative The alternative with the highest sum of values wins Independent of irrelevant alternatives (one intuition: one can assign a value to an alternative without changing the value of other alternatives) More information about range voting available at www.rangevoting.org These still fall prey to strategic voting (e.g., Gibbard-Satterthwaite impossibility, discussed in the next lecture)